PHYSICS COLLEGE

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?

Answers

Answer 1

Answer:

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^(2) - x^(2)}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_(m)}$ ) is

$v_(m) = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$(v)/(v_(m)) = \frac{\sqrt{A^(2) - x^(2)}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_(m)$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& (1)/(4) = \frac{\sqrt{15^(2) - x^(2)}}{15}\n&or,& A = 14.52~cm$

Related Questions

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?(c) What was his total displacement relative to the air mass?
The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light
Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.
Suppose that you run along three different paths from location A tolocation B. Along which path(s) would your distance traveled be different than your displacement (Question 2) please help
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,1. Which terms (if any) are positive?2. Which terms (if any) are negative?3. Which terms (if any) are zero? b) Determine the energy output by the athlete in SI unit.c) Determine his metabolic power in SI unit.d) Another day he performs the same task in 1.2 s.1. Is the metabolic energy that he expends more, less, or the same?2. Is his metabolic power more, less, or the same?

If a microwave oven produces electromagnetic waves with a frequency of 2.30 ghz, what is their wavelength?

Answers

Answer: wavelength is $<strong> 1.30 * 10^8 nm.</strong>$

The frequency of the microwave is, f = 2.30 GHz.

To Find frequency use the formula:

$c=f$ λ

Where, c is the speed of electromagnetic wave or light. f is the frequency, and λ is the wavelength of light.

Rearranging, $\lambda = (c)/(f)$

Plug in the values,

$\lambdam = (3 * 10^8 m/s)/(2.30 GHz(10^9 Hz)/(1 GHz))=0.130 m(10^9 nm)/(1 m) = 1.30 * 10^8 nm.$

$\lambda = (c)/(\nu) = (3 \cdot 10^8)/(2.3 \cdot 10^9) = (3)/(23) \approx 0.130435 \approx 0.13 \ m.$

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5.42 x 10^4 rad/s. In the process, the bit turns through 1.72 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.42 x 10^4 rad/s, starting from rest?

Answers

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

$\alpha = (w2^(2)-w1^(2) )/(2*(\theta2 - \theta1))$

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²

t=1.097 sec

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

$Tan \theta = (y)/(D)$

where D is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy = |y1max (λ1) − y1max (λ2)|

Δy = $|(D\lambda _1)/(d) - (D\lambda _2)/(d) |$

Δy = $D |(d/20)/(d) - (d/15)/(d) |$

Δy = $D |(1)/(20) - (1)/(15) |$

Δy = $4.90 |(1)/(20)- (1)/(15) |$

Δy = 81.7mm

The separation between these maxima is 81.7 mm

Δy = |y₂max (λ1) − y₂max (λ2)|

Δy = $D|(2(d/20))/(d) - (5(d/15))/(2d) |$

Δy = $4.90|(1)/(10) - (1)/(6) |$

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

Learn more about Wave Interference here:

brainly.com/question/16622714

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Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..

Answers

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

Answer:

i believe its a?

Explanation:

In an inelastic collision, momentum is conserved

A shot-putter exerts an unbalanced force of 128 N on a shot giving it an acceleration of 19m/s2. What is the mass of the shot?

Answers

Answer:

128 is the ans cuz N is also lnown as mass

Explanation:

128

Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

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The normal is a line perpendicular to the reflecting surface at the point of incidence.

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If the_____of a wave increases, its frequency must decrease.a. period b. energy c. amplitude d. velocity